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• 2002 AMC 10A Problems/Problem 13
  ...is dropped (the big triangle and two smaller ones it gets split into) are similar! We take the triangle with hypotenuse <math>20</math>. The ratio of sides b [[Category:Introductory Geometry Problems]]
  3 KB (395 words) - 13:22, 8 November 2021
• 2002 AMC 10A Problems/Problem 20
  ...l, triangles <math>\triangle GAD</math> and <math>\triangle HCD</math> are similar. Hence, <math>\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}</math>. ...l, triangles <math>\triangle GAF</math> and <math>\triangle JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore
  3 KB (439 words) - 22:15, 9 June 2023
• 1972 USAMO Problems/Problem 5
  ...orresponding vertices. From the lemma, pentagon <math>A'B'C'D'E'</math> is similar to pentagon <math>ABCDE</math> with <math>AB = m(A'B')</math> and parallelo ...lograms, <math>BD' = EC' = (m - 1)a</math>. Triangle <math>EB'C'</math> is similar to triangle <math>ECB</math>, so <math>me = BC = \left(\frac{2m - 1}{m - 1}
  4 KB (684 words) - 09:29, 26 March 2023
• 1960 IMO Problems/Problem 7
  ...h>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}</mat [[Category:Olympiad Geometry Problems]]
  2 KB (410 words) - 15:25, 23 March 2020
• 2000 AMC 10 Problems/Problem 5
  (a) Triangles <math>ABP</math> and <math>MNP</math> are similar, and since <math>PM=\frac{1}{2}AP</math>, <math>MN=\frac{1}{2}AB</math>. [[Category:Introductory Geometry Problems]]
  1 KB (242 words) - 01:24, 27 July 2023
• 2000 AMC 10 Problems/Problem 16
  Extend line <math>\overline{DC}</math> as above. This creates two similar triangles whose side lengths have the ratio <math>5:4</math>. Therefore <m [[Category:Introductory Geometry Problems]]
  7 KB (966 words) - 23:22, 31 July 2023
• 1995 USAMO Problems/Problem 3
  ...cated on the ray <math>OA_1</math> so that <math>\triangle OAA_1</math> is similar to <math>\triangle OA_2A</math>. Points <math>B_2</math> and <math>C_2</ma [[Category:Olympiad Geometry Problems]]
  2 KB (364 words) - 01:42, 19 April 2024
• 2004 AMC 10B Problems/Problem 18
  ...he triangles <math>\triangle AFG</math> and <math>\triangle AEC</math> are similar, as they have all angles equal. Their ratio is <math>\frac {AF}{AE} = \frac ==Solution 3 (Coordinate Geometry)==
  6 KB (904 words) - 12:54, 22 October 2023
• 2009 AMC 12A Problems/Problem 20
  [[Category: Introductory Geometry Problems]] ...B = \angle DEC</math>, triangles <math>AEB</math> and <math>DEC</math> are similar. Their ratio is <math>\frac {AB}{CD} = \frac {3}{4}</math>. Since <math>AE
  3 KB (543 words) - 21:09, 23 October 2023
• 2009 AMC 10A Problems/Problem 17
  [[Category: Introductory Geometry Problems]] Triangle <math>EAB</math> is similar to <math>BAD</math>, as they have the same angles. Segment <math>BA</math>
  4 KB (684 words) - 21:14, 23 October 2023
• 2009 AMC 10B Problems/Problem 18
  == Solution 1 (Coordinate Geometry)== ...le, thus all their angles are equal, and therefore these two triangles are similar.
  6 KB (930 words) - 22:14, 18 January 2024
• 2009 AMC 10B Problems/Problem 22
  [[Category: Introductory Geometry Problems]] ...and <math>\triangle NQO</math> have the same angles and therefore they are similar. The ratio of their sides is <math>\frac{QM}{OQ} = \frac{\sqrt 5}1 = \sqrt
  12 KB (1,868 words) - 03:36, 30 September 2023
• 2009 AIME I Problems/Problem 4
  ...2E_3}</math> where the points are in order from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2 ==Solution 6(Coordinate Geometry)==
  7 KB (1,117 words) - 00:23, 9 January 2023
• 2009 AIME I Problems/Problem 5
  ...which implies <math>AM||LP</math> and <math>\bigtriangleup{AMB}</math> is similar to <math>\bigtriangleup{LPB}</math> [[Category:Intermediate Geometry Problems]]
  8 KB (1,224 words) - 19:52, 7 March 2024
• 2009 AIME I Problems/Problem 12
  ...ath>AD=\dfrac{12^{2}}{37}</math> and <math>DB=\dfrac{35^{2}}{37}</math> by similar triangles. Let <math>O</math> be the center of <math>\omega</math>; notice Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD
  12 KB (1,970 words) - 22:53, 22 January 2024
• Radical axis
  ...ga_1</math>, for example, we adopt the convention that <math>P = M</math>; similar conventions hold for <math>\omega_2</math>. Power of a Point still holds in [[Category : Geometry]]
  12 KB (2,125 words) - 08:38, 23 May 2024
• 2009 AIME II Problems/Problem 15
  ...>AO = \frac{1}{2}</math>. Since <math>CC'Y</math> and <math>AOY</math> are similar right triangles, we have <math>CY/AY = CC'/AO = \sin(t)</math>, and hence, [[Category: Intermediate Geometry Problems]]
  11 KB (1,862 words) - 21:23, 23 May 2024
• 1965 IMO Problems/Problem 3
  ...vol APWX = k3(k+1)3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k2(k+1)2 times small [[Category:Olympiad Geometry Problems]]
  2 KB (309 words) - 10:52, 30 September 2022
• 2002 AMC 12A Problems/Problem 23
  ...ngle ABD</math> and <math>\triangle ACB</math> are [[Similarity (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>. [[Category:Introductory Geometry Problems]]
  6 KB (899 words) - 01:41, 5 July 2023
• 2010 AMC 12A Problems/Problem 8
  == Solution 3 (Similar Triangles) == ...irc}</math> and <math>\angle ACF=\angle BAE</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2AC</math>, <math>AE=2CF=2FE</math>, as tri
  4 KB (621 words) - 15:12, 21 June 2023

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